Hint on Cohn, Section 1.3, Problem 8 (Measure Theory)

Question

Here's the question:

Suppose $I \subseteq \mathbf{R}$ is a bounded interval, and $B\subseteq I$ satisfies $$\lambda^\ast(I) = \lambda^\ast(B) + \lambda^\ast(B^c \cap I).$$ Show that $B$ is $\lambda^\ast$-measurable.

I'm hoping to get a hint (not a solution) or maybe a suggestion on a simple case to start with.

Note that measurability of $B$ with respect to $\lambda^\ast$ is defined as $$ \lambda^\ast(A) = \lambda^\ast(A \cap B) + \lambda^\ast(A \cap B^c), $$ for all $A \subseteq \mathbf{R}$.

Answer 1

I have a solution (outlined below), but it is more equation manipulation than anything else. If someone can give intuition, I'd be grateful.

$\newcommand{\R}{\mathbf{R}}$ I will assume the following fact.

$E \subseteq \R$ is $\lambda^\ast$-measurable if and only if $\lambda^\ast(J) \geq \lambda^\ast(J \cap E) + \lambda^\ast(J \cap E^c)$ for all open intervals $J \subseteq \R$.

Let $J \subseteq \R$ be an open interval. Since $J$ is $\lambda^\ast$ measurable, it follows that for any $S \subseteq \R$, \begin{equation}\label{eon:meas} \lambda^\ast(S) = \lambda^\ast(J \cap S) + \lambda^\ast(J^c \cap S). \qquad (1) \end{equation}

By assumption, $B\subseteq \R$ satisfies \begin{align*} \lambda^\ast(I) &= \lambda^\ast(B) + \lambda^\ast(I \cap B^c) & \\ &= (\lambda^\ast(J \cap B) + \lambda^\ast(J^c \cap B)) + (\lambda^\ast(J \cap I \cap B^c) + \lambda^\ast(J^c \cap I \cap B^c)) & \text{eq. (1)}\\ &= (\lambda^\ast(J \cap B) + \lambda^\ast(J \cap I \cap B^c)) + ( \lambda^\ast(J^c \cap B)) +\lambda^\ast(J^c \cap I \cap B^c)) & \text{arrange by $J, J^c$}\\ &\overset{(*)}{\geq} \lambda^\ast(J \cap I) + \lambda^\ast(J^c \cap I) & \text{cntble subadd.}\\ &= \lambda^\ast(I) & \text{$\lambda^\ast$-meas. of $J$} \end{align*}

Thus the inequality above is equality. Notice that if $a + b = c + d$, and $a \geq c$ and $b \geq d$, then $$ 0 \leq a - c = d - b \leq 0, $$ so $a = c$ and $b = d$. Applying this to ineq. $(*)$, we've shown for any interval $J \subseteq \R$, $$ \lambda(J \cap I) = \lambda^\ast(J \cap B) + \lambda^\ast(J \cap I \cap B^c). \qquad \text{(2)} $$ Suppose $J \subseteq I$. Then the equation above reduces to $$ \lambda^\ast(J) = \lambda^\ast(J \cap B) + \lambda^\ast(J \cap B^c). $$ On the other hand, suppose that $J \not \subseteq I$, then since $I$ is $\lambda^\ast$-measurable, \begin{align*} \lambda^\ast(J) &= \lambda^\ast(J \cap I) + \lambda^\ast(J \cap I^c) & \text{$\lambda^\ast$-meas of $I$}\\ &= \lambda^\ast(J \cap B) + (\lambda^\ast(J \cap I \cap B^c) + \lambda^\ast(J \cap I^c)) & \text{eq. (2)} \\ &\geq \lambda^\ast(J \cap B) + \lambda^\ast(J \cap B^c) & \text{subadd., $B^c = (I \cap B^c) \cup I^c$}. \end{align*} Thus, we've shown that $B \subseteq \R$ satisfies $$ \lambda^\ast(J) \geq \lambda^\ast(J \cap B) + \lambda^\ast(J \cap B^c), $$ where $J \subseteq \R$ was an arbitrary open interval. The claim follows by the fact.